Key Concepts and Formulas

• Tautology: A statement that is always true, regardless of the truth values of its components.
• Implication: $p \to q$ is equivalent to $\sim p \vee q$.
• De Morgan's Laws:
  • $\sim (p \wedge q) \equiv \sim p \vee \sim q$
  • $\sim (p \vee q) \equiv \sim p \wedge \sim q$
• Law of Excluded Middle: $p \vee \sim p \equiv T$ (True)

Step-by-Step Solution

Step 1: Assume $\odot$ is $\vee$ and $\otimes$ is $\to$ We want to check if the given expression is a tautology with these assignments: $(p \wedge q) \odot (p \otimes q) = (p \wedge q) \vee (p \to q)$

Step 2: Rewrite the implication Using the implication equivalence $p \to q \equiv \sim p \vee q$, we can rewrite the expression: $(p \wedge q) \vee (\sim p \vee q)$

Step 3: Apply associativity and commutativity Rearrange the terms to group similar variables: $(\sim p \vee q) \vee (p \wedge q) \equiv \sim p \vee [q \vee (p \wedge q)]$

Step 4: Simplify using the absorption law Note that $q \vee (p \wedge q) \equiv q$. Therefore, we have $\sim p \vee q$ This is NOT a tautology, so this combination of $\odot$ and $\otimes$ won't work. Our assumption in Step 1 was incorrect.

Step 5: Assume $\odot$ is $\to$ and $\otimes$ is $\to$ We want to check if the given expression is a tautology with these assignments: $(p \wedge q) \odot (p \otimes q) = (p \wedge q) \to (p \to q)$

Step 6: Rewrite the implications Using the implication equivalence $p \to q \equiv \sim p \vee q$, we can rewrite the expression: $(p \wedge q) \to (\sim p \vee q)$

Step 7: Rewrite the implication as a disjunction Using the implication equivalence again, we get: $\sim (p \wedge q) \vee (\sim p \vee q)$

Step 8: Apply De Morgan's Law Apply De Morgan's law to $\sim (p \wedge q)$: $(\sim p \vee \sim q) \vee (\sim p \vee q)$

Step 9: Rearrange terms Use the associative and commutative properties of $\vee$: $\sim p \vee (\sim q \vee q)$

Step 10: Apply the Law of Excluded Middle Since $\sim q \vee q$ is always true (Law of Excluded Middle), we have: $\sim p \vee T$

Step 11: Simplify Since anything ORed with true is true, the expression simplifies to: $T$

Step 12: Conclude that the expression is a tautology Since the expression is always true, it is a tautology. Therefore, $\odot$ corresponds to $\to$ and $\otimes$ corresponds to $\to$.

Step 13: Assume $\odot$ is $\wedge$ and $\otimes$ is $\vee$ We want to check if the given expression is a tautology with these assignments: $(p \wedge q) \odot (p \otimes q) = (p \wedge q) \wedge (p \vee q)$

Step 14: Expand and simplify $(p \wedge q) \wedge (p \vee q) \equiv (p \wedge q \wedge p) \vee (p \wedge q \wedge q) \equiv (p \wedge q) \vee (p \wedge q) \equiv p \wedge q$

This expression is only true when both $p$ and $q$ are true, so it is NOT a tautology.

Step 15: Assume $\odot$ is $\wedge$ and $\otimes$ is $\to$ We want to check if the given expression is a tautology with these assignments: $(p \wedge q) \odot (p \otimes q) = (p \wedge q) \wedge (p \to q)$

Step 16: Rewrite the implication $(p \wedge q) \wedge (\sim p \vee q)$

Step 17: Distribute $(p \wedge q \wedge \sim p) \vee (p \wedge q \wedge q) \equiv F \vee (p \wedge q) \equiv p \wedge q$ This is only true when both $p$ and $q$ are true, so it is NOT a tautology.

Common Mistakes & Tips

• Remember the truth tables and equivalences for logical operators.
• De Morgan's Laws are crucial for simplifying expressions with negations.
• When checking for tautologies, try to simplify the expression as much as possible.

Summary

We tested each of the options by substituting the logical operators into the given Boolean expression and simplifying. We found that when $\odot$ is implication ($\to$) and $\otimes$ is implication ($\to$), the expression simplifies to a tautology (always true). Therefore, the correct assignment is $\odot$ as $\to$ and $\otimes$ as $\to$.

The final answer is \boxed{\to, \to}, which corresponds to option (B).